數學級數和問題

Question

請問
(1+2+3)+(1+2+3+...+7+8)+(1+2+3+...+15)+...+(1+2+...+(n^2-1)

如何計算
n^2 等於 n平方
3,8,15等於 2平方- 1, 3平方-1 , 4平方-1 依此類推

我問的是
"(1 2 3) (1 2 3 ... 7 8) (1 2 3 ... 15) ... (1 2 ... (n^2-1))"

該如何計算喔

"(1 2 3) (1 2 3 ... 7 8) (1 2 3 ... 15) ... (1 2 ... (n^2-1))"

Answer 1

以下Σ的k的範圍都是1到n，符號輸入不方便，省略之。
我們有公式：
Σk=n(n+1)/2
Σk2=n(n+1)(2n+1)/6
Σk4=n(n+1)(2n+1)(3n2+3n-1)/30
本題單獨一項是
1+2+3...+(n2-1)=(n2-1)*n2/2=(n4-n2)/2
n=1時，該項是0，加之不改總和，
所以本題雖然k=2到n，但總和與k=1到n一樣。
故總和
=(1/2)(Σk4-Σk2)
=(1/2)[n(n+1)(2n+1)(3n2+3n-1)/30-n(n+1)(2n+1)/6]
=(1/2)*n(n+1)(2n+1)(3n2+3n-6)/30
=(1/2)*n(n+1)(2n+1)(n2+n-2)/10
=(n-1)n(n+1)(n+2)(2n+1)/20

Answer 2

sigma(1+2+...+(n2-1)=sigma(1+n2)/2
=sigma(1/2)+sigma(n2)/2
=(1+n)/4+n*(n+1)*(2n+1)/12
=(1+n)[3+n(2n+1)]/6
=(1+n)(n2+2)/6